Real Roots and Differentiation

It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $ f(x) := x^4-1102x^3-2015$, we are done.

Observe, $f(0) < 0$ and $f(-2) > 0 $, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.

Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$

It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.